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A compound of element A and oxygen has a mole ratio of A:O = 2:3. If 8.0 grams of the...
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Total mass of oxide = 8.0
Mass of O in the oxide = 2.4
Mass of A in the oxide = 8.0 - 2.4 = 5.6
Moles of O in the oxide:
`mol es = (mass)/(molar mass)`
`= (2.4 grams)/(16.0 (gram)/(mol e)) = 0.15 mol es`
Remember that the ratio of A with O is 2:3 therefore:
2:3 and ?:0.15
`= 0.15 * 2/3 = 0.1 mol es A`
Molar mass of A:
`mol es A = (mass A)/(molar mass A)`
`molar mass A= (mass A)/(mol es A)`
`molar mass A= (5.6)/(0.1)`
`molar mass A = 56 (g)/(mol e)`
Molar mass of the Oxide:
`=56(2) + 16(3)`
`= 160 (g)/(mol e)`
Posted by jerichorayel on April 25, 2013 at 3:06 PM (Answer #1)
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